I know that given $\delta < 1$, we can find $\epsilon < 1$ such that $3^{n^\delta} \in \mathcal O(2^{n^\epsilon})$. However, the definition of approximation factor is not in asymptotics. We should get $3^{n^\delta} \leq 2^{n^\epsilon}$ if we want to say the two theorems are actually same.